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There is an essential qualification to the printed hypothesis. If compactly generated group means merely generated algebraically by a compact subset, the claim is false. In the additive group $\mathbb{Q}$ with its usual topology,is compact, being a convergent sequence together with its limit, and generates every rational number since any integer denominator divides some factorial. The group is metrizable. However, an invariant Radon measure has a common mass $c$ on each singleton. If $c>0$, the compact set $K$ has infinite measure, contradicting finiteness on compact sets. If $c=0$, countable additivity makes the entire countable group have measure zero. Thus it has no nonzero Haar measure. This proves that compact generation does not imply local compactness and disproves the literal unrestricted statement.

The intended existence statement is valid when local compactness is included, either explicitly or in the convention that the group has a compact generating neighbourhood. Here is a construction for a locally compact, compactly generated, metrizable Hausdorff group. No invariant measure is assumed during its construction.

Enlarge a compact generating set to a symmetric compact neighbourhood of the identity element $S$. Then $G={\bigcup }_n{S}^n$, and each $S^n$ is compact. Moreover $S^n$ lies in the interior of $S^{n+1}$, since multiplication by the open neighbourhood of the identity element contained in $S$ supplies an open neighbourhood of every point of $S^n$. This gives a compact exhaustion. Every compact metric space has a countable dense subset: take finite $1/k$-nets and unite them. Their union over this exhaustion is dense in $G$, so the metric space is separable and has a countable open basis.

We use nonnegative compactly supported continuous functions and cutoffs equal to one on specified compact sets. Their existence is covered by the source's permission concerning auxiliary functions. Fix a nonzero such function $h$. For nonzero $u\ge 0$ define the weighted covering quantityThe quantities are finite: an open set where $u$ is bounded below has translates covering the compact support of $f$, so finitely many sufficiently large weighted translates dominate it. The denominator is positive, since evaluation at a point where $h>0$ gives $(h:u)\ge h(y)/\Vert u{\Vert }_{\infty }>0$.

Each $I_u$ is nonnegative, homogeneous, monotone, left invariant and subadditive; these follow respectively by rescaling, comparing or combining covers, and translating their centers. Also $I_u(h)=1$. Composing a cover of $f$ by translates of $h$ with a cover of $h$ by translates of $u$ givesThis is a bound independent of $u$. On functions supported in a fixed compact set $C$, choose a cutoff $k\ge 1$ on $C$. Monotonicity and subadditivity implyThus the covering functionals are uniformly Lipschitz on each fixed support.

Choose a nonzero $u_n\ge 0$ with support in an neighbourhood of the identity element shrinking to $e$. The nonnegative functions supported in each compact member of the exhaustion form a separable metric cone in the supremum norm, as a subspace of continuous functions on that compact metric space. Select a countable dense family from each cone and include $h$. The uniform scalar bounds permit a diagonal subsequence along which $I_{u_n}$ converges on their countable union. The displayed Lipschitz estimate extends convergence to every nonnegative member of $C_c(G)$. Denote the limit by $I$. It inherits positivity, homogeneity, monotonicity, subadditivity, invariance and $I(h)=1$.

The crucial remaining step is additivity. Fix $f_1,f_2\ge 0$ with combined support $C$, and fix a compact neighbourhood of the identity element $V_0$. Choose a cutoff $k$ equal to one on $C{V}_0^{-1}{V}_0$. For $\delta >0$, putsetting the ratio to zero off the denominator's nonzero set. These ratios are continuous: $k=1$ on a neighbourhood of the supports, and off those supports the numerator vanishes. Their sum is at most one. Compactness and continuity give, for each $\eta >0$, an neighbourhood of the identity element $V$ small enough that each $r_i$ oscillates by at most $\eta$ on any translate of $V$ meeting $C$. To justify this uniform choice, such translates lie inside the fixed compact set $C{V}_0^{-1}{V}_0$ when $V\subset V_0$; continuity of $(y,v)\mapsto r_i(yv)$ and a finite cover of that compact set give a single small right-increment neighbourhood, then choose $V^{-1}V$ inside it.

If $u$ has support in $V$ and $F\le {\sum }_j{c}_j{T}_{x_j}u$, discard translates not meeting $C$ when covering $f_i$. Let $a_{ij}$ be the supremum of $r_i$ on the remaining translate's support. Then $f_i\le {\sum }_j{a}_{ij}{c}_j{T}_{x_j}u$, and comparing both suprema with the ratios at one common point gives $a_{1j}+a_{2j}\le 1+2\eta$. Taking covering infima yieldsAlong the subsequence this holds eventually, since its supports shrink. Pass to $I$, then let $\eta \downarrow 0$ and $\delta \downarrow 0$; the uniform bound on $I_u(k)$ gives $I(f_1)+I(f_2)\le I(f_1+f_2)$. Subadditivity gives the reverse inequality. Thus $I$ is additive.

Extend $I$ linearly to real functions using positive and negative parts, and then complex-linearly to $C_c(G)$. The compact-support bounds above make it a positive, locally bounded functional. The Riesz representation on compactly supported continuous functions gives a Radon measure $\mu$ with $I(f)=\int fd\mu$. Translation invariance of $I$ and uniqueness in that representation give $\mu (xE)=\mu (E)$ for Borel $E$. It is nonzero because $I(h)=1$, and finite on compact sets because a cutoff dominating their indicators has finite integral. Finally every nonempty open set has positive measure: otherwise its translates cover the compact support of $h$, a finite subcover would give that support measure zero, contradicting $I(h)=1$. Hence $\mu$ is a Haar measure under the stated local-compactness qualification. This is Haar measure from normalized covering functionals, with the limiting additivity proved explicitly.

Write the two formal pseudodifferential operators in normal order aswhere $p={\alpha }_1$ and $q={\alpha }_2$. The formal pseudodifferential composition rule givesThus $\boxed{\mathrm{ord}(P_1{P}_2)=p+q}$ whenever $ab$ is not identically zero, in particular when both leading coefficients are nowhere zero. In general the rigorous statement over the specified coefficient algebra is $\mathrm{ord}(P_1{P}_2)\le p+q$.

The leading scalar coefficients commute. Subtracting the reversed composition also cancels the terms involving $a_1,b_1$, leavingTherefore the order filtration of formal pseudodifferential operators giveswith equality precisely when the displayed coefficient is not identically zero. The bound can be strict: constant-coefficient powers of $\partial$ commute, giving the zero operator, whose order can be assigned $-\infty$.

There is also a genuine qualification to product-order addition in $C^{\infty }(\mathbb{R})$. Choose nonzero smooth functions $a,b$ with disjoint compact supports and take $P_1=m_a,P_2=m_b$. Each has order zero, but their product is zero. Nonzero coefficients as elements of this algebra need not have a nonzero product. Thus an unqualified assertion of exact product order requires an additional leading-symbol hypothesis.

Put $r=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/r$, and take a compactly supported Lighthill stress tensor. The isotropic term in that tensor is $(p-c_0^2\rho ){\delta }_{ij}$: the Kronecker delta is implicit in the printed shorthand. Convolving the Lighthill acoustic analogy with the retarded acoustic Green function and integrating twice by parts in the source coordinates givesThe boundary terms vanish because of compact support. For source size $\ell$, the acoustic compact-source approximation requires $\omega \ell /c_0\ll 1$. At $r\gg \ell$, we may therefore replace the denominator by $r$ and the retarded argument by $t-r/c_0$ in the leading source integral. Define $S_{ij}(t)=\int T_{ij}(\mathbf{y},t)d^{3}y$. In the radiation region $\omega r/c_0\gg 1$, the leading two spatial derivatives act on the retarded time rather than on $1/r$ or $\mathbf{n}$:Thus the leading acoustic quadrupole density isFor the compact acoustic quadrupole Mach-number scaling, let $U$ be the source fluctuation speed, $m=U/c_0\ll 1$, and use the advective source time $\ell /U$. With $T_{ij}=O({\rho }_0{U}^2)$, one has $S_{ij}=O({\rho }_0{U}^2{\ell }^3)$ and ${\ddot{S}}_{ij}=O({\rho }_0{U}^4\ell )$. ConsequentlyThe fourth power refers to the radiation coefficient with geometric spreading separated. It assumes that the source frequency scales as $U/\ell$; an independently imposed frequency or an independently strong thermal source would change that Mach number estimate.

Let $I^{ij}(t)=\int T^{00}(t,\mathbf{x})x^i{x}^j{d}^{3}x$, using spatial indices $i,j$. Stress-energy conservation, symmetry of the stress-energy tensor and integration by parts giveand a second derivative givesAll boundary terms vanish by compact support. This proves the stress-energy quadrupole identity,The support is understood to remain bounded over the time interval used for differentiating these integrals.

For the radiation calculation set $c=1$, write $g_{ab}={\eta }_{ab}+h_{ab}$ and use the trace-reversed metric perturbation ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$ in Lorenz gauge in linearized gravity. With the curvature and field-equation signs in this paper, the linear equation isThe retarded solution with no incoming radiation is consequentlyLet $r=|\mathbf{x}|$ be the distance from the source's centre of mass, and let $d$ be its size. In the radiation zone, $r\gg d$. The leading multipole expansion additionally assumes slow internal motion, or $d$ small compared with the radiation wavelength. These assumptions allow $|\mathbf{x}-{\mathbf{x}}^{\prime }|\simeq r$ in the denominator and the common retarded time $u=t-r$ at leading quadrupole order. ThusDefine the trace-free mass quadrupole moment byTo make the requested normalization explicit, write the trace-free spatial metric coefficient as $h_{ij}^{\mathrm{TF}}=-2E_{ij}$. Trace reversal only adds a spatial trace, so the preceding equations give the far-zone quadrupole strain coefficientThe supplied scan does not contain the referenced perturbation handout. The definition above fixes the normalization of $E$ directly. If $E$ denotes the physical transverse-traceless tensor in that handout, the right-hand side is also to be transversely projected for the observation direction. Trace-free alone does not imply transverse. With $\mathbf{n}=\mathbf{x}/r$, putThe transverse-traceless projector gives $E_{ij}^{\mathrm{TT}}={\Lambda }_{ij,kl}{E}_{kl}$ and the physical spatial strain is $H_{ij}^{\mathrm{TT}}=2E_{ij}^{\mathrm{TT}}$. This also states the result without relying on omitted handout notation. The retarded construction and direction-dependent transverse projection are described in Carroll's general relativity lectures, with the appropriate changes of metric-sign convention.

For the equal-mass binary star, its centre of mass is at the origin and the separation is $2R$. The gravitational force on either star is $G{m}^2/(2R{)}^2$, while its centripetal force is $mR{\omega }^2$. HenceAt leading Newtonian order, $T^{00}$ is the sum of the two point-mass densities. Its moments areso the trace-free mass quadrupole moment isThe Newtonian bound system is treated in the usual leading weak-field quadrupole approximation; an exactly conserved effective source also includes the interaction stresses. Accelerated point-mass dust alone would not satisfy the conservation hypothesis.

Differentiating twice and evaluating at $u=t-r$ givesFor an observer on the $z$ axis this is already transverse and traceless. Its alternating plus polarization and cross polarization have a phase difference of $\pi /2$, giving circularly polarized gravitational radiation at twice the orbital frequency. The amplitude decays as $1/r$, as expected for outgoing radiation in three spatial dimensions.

More generally take the line of sight in the $xz$ plane at inclination $\iota$ to the orbital angular-momentum axis. In the transverse basis ${\mathbf{e}}_{\theta }=(\cos \iota ,0,-\sin \iota )$, ${\mathbf{e}}_{\varphi }=(0,1,0)$, define $A=4Gm{R}^2{\omega }^2/r$. Projection gives the physical gravitational wave polarizationsThus a generic view is elliptically polarized, the face-on view is circularly polarized and the edge-on view is linearly polarized. This is circular-binary quadrupole radiation; an overall sign or transverse-axis rotation only changes the polarization convention.

There is a definitional qualification. If a compactly generated group means a group algebraically generated by a compact subset, the assertion as printed is false. In the usual topology on the additive group $\mathbb{Q}$, the setis compact and generates $\mathbb{Q}$: every positive integer denominator divides some factorial. The group is metrizable, but it has no nonzero Haar measure finite on compact sets. Indeed, translation invariance gives a common singleton mass $a\ge 0$. If $a>0$, the infinite compact set $K$ has infinite mass; if $a=0$, countable additivity gives zero mass to the whole countable group. This proves that compact generation does not imply local compactness or existence of Haar measure.

If the convention includes a compact generating neighbourhood, local compactness is already part of the hypothesis. More explicitly, the standard corrected result is every locally compact, compactly generated metrizable group has Haar measure. Here is a construction, rather than an invocation of that existence theorem.

Fix a nonzero $h\ge 0$ in the space of continuous compactly supported functions $C_c(G)$. For nonzero $u\ge 0$ in $C_c(G)$ and $f\ge 0$ in $C_c(G)$, define the covering number and its normalized functional byThese numbers are finite: translates of the region where $u>0$ cover the compact support of $f$, and a finite subcover, multiplied by a sufficiently large coefficient, dominates $f$. They are positive when $f\ne 0$, since every cover has total coefficient at least $\Vert f{\Vert }_{\infty }/\Vert u{\Vert }_{\infty }$. Composition of covers gives $(f:u)\le (f:h)(h:u)$. Thus, with $M_f=(f:h)$,The functionals are homogeneous, monotone, subadditive and exactly invariant under left translation; the last property follows by relabelling the covering translates.

The key step in Haar measure from normalized covering functionals is asymptotic additivity as $\mathrm{supp}u$ shrinks to $e$. Given $f_1,f_2\ge 0$, put $f=f_1+f_2$ and choose $p\in C_c(G)$ with $p\ge 1$ on both supports. For $\delta >0$, set ${\alpha }_i=f_i/(f+\delta p)$, extended by zero where the denominator vanishes. These are continuous compactly supported functions and ${\alpha }_1+{\alpha }_2\le 1$. For any $\eta >0$, a sufficiently small identity neighbourhood $V$ ensuresTake $\mathrm{supp}u\subseteq V$ and any cover $f\le {\sum }_j{a}_j{T}_{x_j}u$, where $T_{x}u(t)=u(x^{-1}t)$. Since $f_i\le {\alpha }_{i}f+\delta p$, it follows thatAdding and then taking the infimum over the cover of $f$ givesFirst make $\delta$ small, then $\eta$ small, and finally the support of $u$ small enough for the oscillation bound. The additivity defect therefore tends to zero.

Compact generation gives a countable compact cover by finite products of a compact symmetric generating set. Local compactness upgrades this to a compact exhaustion with interiors covering $G$. Metrizability supplies a countable identity-neighbourhood base and a countable family of nonnegative $C_c$ functions dense in the uniform norm on each fixed compact support; this is the countable family whose existence may be assumed. Choose nonzero $u_n\ge 0$ with supports shrinking to $e$. The uniform bounds $I_{u_n}(f)\le M_f$ and a diagonal subsequence give convergence on that countable family. To extend convergence to every nonnegative $f$ supported in a fixed compact $L$, choose a cutoff $p_L\ge 1$ on $L$. Monotonicity and subadditivity givefor $f,g$ supported in $L$. Uniform approximation therefore makes the chosen subsequence Cauchy on every such $f$.

Let $I(f)$ be its limit. Asymptotic additivity, homogeneity and invariance give an additive positive invariant functional on the nonnegative cone, with $I(h)=1$. Extend it by differences to real $C_c(G)$ and then complex linearly. It is bounded on each fixed compact support by the cutoff bound. The Riesz representation on compactly supported continuous functions consequently produces a positive Radon measure $m$ with $I(f)=\int fdm$. Invariance of $I$ and uniqueness of that representation imply $m(xE)=m(E)$. The measure is nonzero because $I(h)=1$, and finite on compact sets by the cutoff estimate. Moreover, for any nonzero $f\ge 0$, composition of covers gives $I_u(f)\ge 1/(h:f)>0$, so every nonempty open set has positive measure. HenceThis completes the intended construction under the convention supplying local compactness, while the rational example settles the weaker literal interpretation.

Fix Haar measure $m_G$ and use $\hat{f}(\chi )={\int }_{G}f(x)\overline{\chi (x)}d{m}_G(x)$. A positive-definite function $p$ satisfies ${\sum }_{i,j}{z}_i\overline{z_j}p(x_i-x_j)\ge 0$ for every finite choice of points and coefficients. The Bochner theorem states that a continuous positive-definite function on a locally compact Hausdorff Abelian group has a unique representationwhere ${\mu }_p$ is a Radon measure. Conversely, any finite positive Radon measure on the Pontryagin dual group has a continuous positive-definite inverse transform. The sign convention here is the one compatible with the displayed Fourier transform.

First note that the dual is a locally compact Hausdorff topological group. Its group operations are continuous in the compact-open topology proved in question 3. For local compactness, adjoin the zero functional to the character space of an algebra of $L^1(G)$. All its elements lie in the closed unit ball of $L^1(G{)}^{\ast }$, and the equations $\phi (a\ast b)=\phi (a)\phi (b)$ define a closed subset in the weak-star topology. The Banach-Alaoglu theorem makes that subset compact Hausdorff. Removing the zero functional gives an open, hence locally compact, subspace. This provides the setting for constructing the measure on the dual, without assuming a Fourier inversion theorem.

Let $u^{\ast }(x)=\overline{u(-x)}$ and let $\mathcal{P}$ be the finite sums of convolution squares $u\ast u^{\ast }$, $u\in C_c(G)$. Each such square is continuous with compact support, and is positive-definite becauseIts transform is $|\hat{u}{|}^2$, so every $p\in \mathcal{P}$ has $\hat{p}\ge 0$. Also $p\ast q\in \mathcal{P}$: convolution of two squares is $(u\ast v)\ast (u\ast v{)}^{\ast }$, and finite sums distribute. Insert the Bochner representation of $q$ into $p\ast q$ and use the Fubini theorem to obtainThe representing measure on the right is finite and positive. Uniqueness in the Bochner theorem and commutativity now give the Bochner consistency identity for convolution squares

For every compact $K\subseteq \hat{G}$ there is $p\in \mathcal{P}$ with $\hat{p}>0$ throughout $K$. Indeed, at any ${\chi }_0$ choose $u={\chi }_{0}a$, where $a\ge 0$ is a nonzero member of $C_c(G)$. Then $\hat{u}({\chi }_0)=\int a>0$, and continuity supplies a neighbourhood where it stays nonzero. A finite subcover of $K$ and the sum of the corresponding squares give such a $p$.

For $\psi \in C_c(\hat{G})$, choose $p$ positive in transform on its support and defineThe quotient is set to zero away from the support; its denominator is bounded away from zero there. If $q$ is another choice, the consistency identity, multiplied by $\psi /(\hat{p}\hat{q})$ on that compact support, shows that the integral is unchanged. Choosing one $p$ for a union of supports proves linearity; positivity follows from positivity of ${\mu }_p$. On each fixed compact support it is bounded by ${\mu }_p(K)/{\min }_K\hat{p}$ times the uniform norm. Moreover, for every $p\in \mathcal{P}$ and $\psi \in C_c(\hat{G})$, choosing $q$ positive on the support of $\psi$ givesA nonzero square has ${\mu }_p(\hat{G})=p(0)=\Vert u{\Vert }_2^2>0$; a suitable nonnegative compactly supported $\psi$ consequently makes this last expression positive. Thus $J$ is nonzero.

To prove translation invariance, fix $\eta \in \hat{G}$ and modulate $p$ by $p_{\eta }(x)=\eta (x)p(x)$. This again belongs to $\mathcal{P}$, since modulation takes $u\ast u^{\ast }$ to $(\eta u)\ast (\eta u{)}^{\ast }$. Direct calculation and Bochner uniqueness giveHence, for ${\psi }_{\eta }(\chi )=\psi ({\eta }^{-1}\chi )$, using $p_{\eta }$ as the denominator function and changing variables gives $J({\psi }_{\eta })=J(\psi )$. The Riesz representation on compactly supported continuous functions now produces a nonzero translation-invariant positive Radon measure $\nu$ on the dual. It is finite on compact sets, so it is Haar measure. This is the compatible dual Haar measure, denoted $m_{\hat{G}}=\nu$.

The identity for $J(\psi \hat{p})$ says, by uniqueness of Radon measures,Thus $\hat{p}\in L^1(\hat{G},\nu )$ and the Bochner representation immediately provesIt also proves the formula for all finite complex linear combinations of these functions. The measure scale is fixed by this identity: at $x=0$, a nonzero square gives $\int |\hat{u}{|}^{2}d\nu =\Vert u{\Vert }_2^2$.

A wider inversion class is $f\in L^1(G)$ with $\hat{f}\in L^1(\hat{G},\nu )$. To prove this extension, choose nonnegative $a_U\in C_c(G)$ with integral one and support in a small identity neighbourhood $U$, and put $p_U=a_U\ast a_U^{\ast }$. These functions form an approximate identity: they are nonnegative, have integral one and support in $U-U$. Also $0\le {\hat{p}}_U=|{\hat{a}}_U{|}^2\le 1$, and ${\hat{p}}_U\to 1$ uniformly on compact subsets of the dual. For this uniform convergence use equicontinuity of compact families of characters: evaluation is jointly continuous, as follows from the translation quotient in question 3, and a finite cover of a compact family gives uniform control near the identity.

Convolving the already established inversion formula for $p_U$ with $f$, the Fubini theorem givesThis is justified by $\Vert f{\Vert }_1{\mu }_{p_U}(\hat{G})<\infty$. Compact-set uniform convergence and an integrable-tail split implyExplicitly, first choose a compact set outside which $\int |\hat{f}|$ is small, use the bound $|{\hat{p}}_U-1|\le 2$ on that tail, and use uniform convergence on the compact remainder. This argument works for neighbourhood-indexed nets, without assuming metrizability. Consequently $f\ast p_U$ converges uniformly to the continuous inverse integral. On the other hand, translation continuity in Lp on a locally compact group gives $f\ast p_U\to f$ in $L^1$; testing these two limits against $C_c(G)$ shows they agree almost everywhere. At every continuity point of the chosen representative of $f$, the shrinking-support approximate identity also converges pointwise to $f(x)$. We obtain Fourier inversion on a locally compact abelian group:valid everywhere for continuous $f\in L^1(G)$ with integrable transform, and in general as equality with a continuous representative almost everywhere. This class includes all convolution squares above, and their linear span.

Finally, for arbitrary $h\in C_c(G)$, apply the inversion formula at zero to $h\ast h^{\ast }$:Thus the integral Fourier transform is linear and norm-preserving on $C_c(G)$. That space is dense in $L^2(G)$, so for $f\in L^2(G)$ choose $h_n\in C_c(G)$ with $h_n\to f$ in $L^2$ and define $\mathcal{F}f$ as the $L^2(\hat{G},\nu )$ limit of ${\hat{h}}_n$. The norm identity makes this limit exist and independent of the approximating sequence, and preserves linearity. This proves the Plancherel theorem for locally compact abelian groups in the requested form:On $L^1\cap L^2$, choose the approximating sequence to converge in both norms; uniform convergence of the integral transforms and their $L^2$ convergence show that this extension agrees almost everywhere with the original transform. For general $L^2$ functions the extension is defined by norm limits, not by presuming that the pointwise integral exists.

The integral identity requires compactness of $M$, compact support of $X$, or suitable conditions eliminating flux at infinity. The printed question does not state such a hypothesis. We prove the Riemannian divergence theorem for a compact manifold, and more generally for a compactly supported smooth vector field on an oriented Riemannian manifold with boundary.

Write $j:\partial M\hookrightarrow M$ for the inclusion. Give $\partial M$ the outward-normal-first boundary orientation, and let $N$ be the outward unit normal vector. The Riemannian volume form of the induced Riemannian metric isIndeed, if $(e_1,\dots ,e_{n-1})$ is a positively oriented orthonormal basis tangent to the boundary, $(N,e_1,\dots ,e_{n-1})$ is a positively oriented orthonormal basis in $M$. Both sides therefore evaluate to one on the boundary basis.

At a boundary point decompose $X=g(X,N)N+X^{\top }$, where $X^{\top }$ is tangent to $\partial M$. Evaluating on $n-1$ tangent vectors, the term with $X^{\top }$ vanishes: all $n$ arguments of the alternating volume form then lie in the $(n-1)$-dimensional boundary tangent space. Thus the interior product of a differential form satisfiesThe general Stokes theorem for an oriented manifold with boundary states ${\int }_{M}d\eta ={\int }_{\partial M}{j}^{\ast }\eta$ for any smooth differential form $\eta$ of degree $n-1$ with compact support; compactness of $M$ makes the support condition automatic. Applying it to $\eta ={\iota }_X{\omega }_g$ and using the defining equation for the divergence of a Riemannian vector field proves

For a concrete failure without the extra hypothesis, take $M=[0,\infty )$ with Riemannian metric $d{x}^2$ and $X=(\arctan x){\partial }_x$. The divergence of a Riemannian vector field is $\mathrm{div}X=(1+x^2{)}^{-1}$, sobecause $X$ vanishes at the only boundary point. Both integrals are finite. The missing $\pi /2$ is the limiting flux at infinity, so mere integrability of the divergence cannot repair the unrestricted assertion.

For a kernel for density estimation $K$ and smoothing bandwidth $h>0$, the kernel density estimator isWe use the convention that a kernel of order ell hasThe kernel for density estimation is integrable and can have negative values; requiring nonnegativity would prevent its second moment from vanishing. Some conventions additionally require a nonzero moment of order $\ell$. The construction below covers that convention too.

The Legendre polynomial kernel construction uses the polynomial reproducing kernel at zero:Every polynomial $p$ of degree at most $\ell -1$ has the expansion $p={\sum }_{m=0}^{\ell -1}⟨p,{\varphi }_m⟩{\varphi }_m$, since the displayed Legendre polynomials have those successive degrees and form an orthonormal basis of this polynomial space. ThereforeTaking $p=1,u,\dots ,u^{\ell -1}$ proves all the required moment identities. Boundedness and compact support give every absolute moment and a finite L2 norm. If a nonzero moment of order $\ell$ is required, replace this kernel byThe lower moments are unchanged by orthogonality. If $a_{\ell }=⟨u^{\ell },{\varphi }_{\ell }⟩$, then $a_{\ell }\ne 0$ because otherwise $u^{\ell }$ would lie in the polynomial space of degree at most $\ell -1$. Evaluating its full Legendre expansion at zero gives $\int u^{\ell }{K}_{\ell }=-{\varphi }_{\ell }(0)a_{\ell }$, so $\int u^{\ell }{\tilde{K}}_{\ell }=a_{\ell }\ne 0$.

For instance, a simple kernel satisfying the moment conditions needed for three derivatives isIts negative wings cancel its central contribution to the second moment.

Here is the third-order pointwise kernel error bound. Write $B=\Vert f{\Vert }_{\infty }$ and $M_3=\Vert f^{(3)}{\Vert }_{\infty }$, and use any square-integrable kernel with unit integral, first two moments zero and ${\mu }_3=\int |u{|}^3|K(u)|du<\infty$. A change of variables givesBy the Taylor theorem with Lagrange remainder,The moment cancellations consequently give the uniform bias of a kernel density estimator boundFor the variance, independence and another change of variables giveThe triangle inequality and Cauchy-Schwarz inequality now yieldBalancing $h^3$ against $(nh{)}^{-1/2}$ givesThe constant is independent of both $n$ and $x$. The original PDF has exponent $3/7$; the converted TeX's $3/2$ is a transcription error.

To obtain the final parametric rate, we choose a flat-top kernel for density estimation. This kernel also satisfies all the preceding moment conditions, so a single kernel works for both constructions, with different bandwidth choices. With the Fourier transform convention $\hat{g}(t)=\int e^{-itx}g(x)dx$, choose a real even $C^{\infty }$ function $\eta$ equal to one on $[-1,1]$ and zero outside $[-2,2]$, and putSuch a cutoff is explicit: if $\rho (s)=e^{-1/s}$ for $s>0$ and zero otherwise, take $\eta (t)=\rho (2-|t|)/[\rho (2-|t|)+\rho (|t|-1)]$. It is constant near zero, and its endpoint transitions are smooth. Repeated integration by parts shows that $K$ and all its derivatives decay faster than every power, so $K$ lies in the Schwartz space. The Fourier inversion and moment differentiation of the Fourier transform giveThus this kernel can be used with $h_n=n^{-1/7}$ for the preceding general bound.

For the sharper example choose the probability density functionTo check this is a density and identify its Fourier transform, direct integration givesThe continuous value at zero is as displayed. This function is nonnegative and integrable, being bounded near zero and $O(x^{-2})$ at infinity. Fourier inversion gives $\hat{f}(t)=(1-|t|{)}_{+}$, and in particular $\int f=\hat{f}(0)=1$. Differentiation of this compact frequency integral also shows that $f$ and $f^{(3)}$ are bounded. Since $\eta =1$ on the support of $\hat{f}$, the convolution theorem givesBoth sides are continuous, so the equality holds at every $x$. There is therefore exactly zero bias of a kernel density estimator at bandwidth one, and the variance bound above givesThis is parametric-rate kernel estimation of a bandlimited density. A fixed bandwidth is appropriate because it reproduces this particular density exactly; a generic density would retain a nonzero smoothing bias.